Question

A point O is in the interior of Δ ABC .Prove that 2(OA +OB +OC )> AB +BC +AC .​

Answer 1

Step-by-step explanation:

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA

*Construction"

BO is produced to intersect AB at D.

For ΔABD

AB+AD>BD

⇒AB+AD>BO+OD.....[1]

For ΔCOD

CD+OD>OC......[2]

Adding [1] and [2] we get

AB+AD+CD+OD>BO+OD+OC

⇒AB+AC+OD>BO+OD+OC

⇒AB+AC>OB+OC.....(3)

Similarly

AB+BC>OA+OC.....(4)

And

AC+BC>OA+OB.....(5)

Adding (3),(4)and (5) we get

2(AC+BC+